(树) 剑指 Offer 32 - I. 从上到下打印二叉树 —

(树) 剑指 Offer 32 - I. 从上到下打印二叉树 ——【Leetcode每日一题】

❓剑指 Offer 32 - I. 从上到下打印二叉树

难度：中等

从上到下打印出二叉树的每个节点，同一层的节点按照从左到右的顺序打印。

例如:
给定二叉树: [3,9,20,null,null,15,7],

[3,9,20,15,7]

提示：

节点总数 <= 100

💡思路：BFS

使用优先队列进行层序遍历即可！

🍁代码：(C++、Java)

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> levelOrder(TreeNode* root) {
        vector<int> ans;
        if(root == nullptr) return ans;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            TreeNode* temp = q.front();
            q.pop();
            ans.push_back(temp->val);
            if(temp->left != nullptr) q.push(temp->left);
            if(temp->right != nullptr) q.push(temp->right);
        }
        return ans;
    }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] levelOrder(TreeNode root) {
        ArrayList<Integer> ans = new ArrayList<>();
        if(root == null) return new int[0];
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        while(!q.isEmpty()){
            TreeNode temp = q.poll();
            ans.add(temp.val);
            if(temp.left != null) q.add(temp.left);
            if(temp.right != null) q.add(temp.right);
        }
        int[] ret = new int[ans.size()];
        for(int i = 0; i < ans.size(); i++){
            ret[i] = ans.get(i);
        }
        return ret;
    }
}